Yes the limit should be negative when approaching from the left. The thing is when you have ##x\sqrt{1+\frac{1}{x^2}} = \sqrt{x^2+1}## that's only true for ##x> 0##.
The second one can be rewritten in two part (if you again take care of the sign). One of them being ##\frac{1}{\sqrt{1+x}}## which has a limit as ##x\to 0##.

You did it correctly. It's just that you have to remember that for
##x> 0## we have ##x\sqrt{1+\frac{1}{x^2}} = \sqrt{x^2+1}## and for
##x < 0## we have ##x \sqrt{1+\frac{1}{x^2}} = -\sqrt{x^2+1}##.
You could also write this as ##x\sqrt{1+\frac{1}{x^2}} = \text{sign} (x) \sqrt{x^2+1}## which always is true.
Because the square-root of a number is always positive, that's the definition.
For example when you solve ##x^2=1## you write ##x=\pm \sqrt{1}## since the ##\sqrt{1}## only refers to the positive root. (Sorry If this wasn't what you were confused about, just guessed it may be.)

You can find the limit from the left by writing a correct form that applies when ##x < 0##, then take the limit. The form (in (1)) that you wrote (with ##\sqrt{x^2+1}##) applies when ##x > 0##. Make a change for ##x < 0## and you are almost done. Alternatively, if ##x < 0## you can write ##x = -t##, where ##t > 0##, then take the limit from the right as ##t \to 0##.